Suppose $(V,\Vert.\Vert)$ is a normed linear space(NLS) and $E$ is an open connected set in $V$,show that $E$ is pathwise connected.
Proof:
Definition:A subset $E$ of a normed linear space $V$ is said to be polygonally connected if for each $x,y\in E$,$\exists x=x_0,x_1,…,x_{n-1},x_n=y$ in $E$ such that $\{tx_{k-1}+(1-t)x_k : t\in [0,1]\}\subset E$
.
In this problem,it suffices to show that $E$ is polygonally connected.Define an equivalence relation on $E$ as follows:
$x\sim y$ iff $x$ and $y$ have a polygonal connection in $E$.
Verify that $\sim$ is an equivalence relation on $E$.
So,$\sim$ generates a quotient set $E/\sim$ ,the set of all equivalence classes of $E$ by $\sim$.
Now we shall show that $E/\sim$ is a singleton and contains $E$ only.
Let,$x\in E$,then $[x]\in E/\sim$
We shall show below that $[x]$ is closed in $E$.
For that we take a convergent sequence $(x_n)$ in $[x]$ which converges to $x_0\in E$.We will show that $x_0\in [x]$.
Since $x_0\in E$ and $E$ is open,so $\exists r>0$ such that $B(x_0,r)\subset E$.
Note that $B(x_0,r)$ is open in $E$.
Now $(x_n)\to x_0$,hence $x_n\in B(x_0,r)$ eventually($\forall n\geq N$ ,say).
Now $V$ is a normed linear space,so $B(x_0,r)$ is convex ,so for an $n\geq N$,$x_n$ and $x_0$ are connected polygonally and hence $x_n\sim x_0$ but $x_n\in [x]$.So,$x_0\in [x]$.
Thus $[x]$ is closed in $E$.
Now we show that $[x]$ is open in $E$.
We shall take a sequence $(a_n)\subset E$ converging to a point $a$ of $[x]$ and show that $a_n\in[x]$ eventually.
since $a\in [x]\subset E$ and $E$ is open,so $B(a,r)\subset E$ for some $r>0$.
Clearly any point of $B(a,r)$ is polygonally connected to $a$ as balls are convex in NLS $V$.
and also $a\in [x]$,so $B(a,r)\subset [x]$.Now $(a_n)\to a$ and hence eventually $(a_n)$ is in $B(a,r)$ and hence $a_n \in [x]$ eventually.
So,$[x]$ is open in $E$.
So,$[x]$ is a non-empty closed-open set in $E$ and $E$ is connected.So,$[x]=E$ and we are done.
So,any two points in $E$ have a polygonal connection between them.So,$E$ is polygonally connected and is obviously path connected.
Is there anything wrong in this proof?Can anyone suggest any alternative way?
Best Answer
Your proof is correct. Here is a shorter proof.
Clearly each path component $P$ of an open subset $E \subset V$ is open in $V$ because if $x \in P$ and $B(x,r) \subset E$, then $B(x,r) \subset P$. Thus the path components of $E$ form a partition of $E$ into pairwise disjoint open subsets.
If $E$ is connected and has more than one path compenent, we get a non-trivial partition into pairwise disjoint open subsets which is a contradiction.